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3 years ago

How do I give the protons, electrons, and neutrons for an isotopes.

1 answer:

3 0

For an isotope of a certain element, only the number of neutrons are changed, the number of protons stays the same because if you change the proton it would mean that it is a different element already. The number of neutrons is calculated from the number shown in the isotopes. That number represents the total protons and neutrons.
<span>
Iodine-127 : protons = 53, neutrons = 74
</span>Iodine-126 : protons = 53, neutrons = 73
<span>Iodine-128 : protons = 53, neutrons = 75</span>

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Which structural feature will differ among an ester, an alcohol, and an aldehyde, regardless of which examples are being compare

AlekseyPX

Answer:

The functional group

Explanation:

A functional group defines a hydrocarbon. an ester the functional group is an O- alkyl group while that one of an alcohol is -OH group. An aldehyde is a functional group on its own.

7 0

4 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

SCORPION-xisa [38]

Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.

Observation and Hypothesis. ...

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Testing and Error Estimation. ...

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3 years ago

Predict the products of the thermal decomposition of strontium nitrate and show the reaction as a word equation.

Schach [20]

yeag

Explanation:

2SrO + 4NO2 + O. The thermal decomposition of strontium nitrate to produce strontium oxide, nitrogen dioxide and oxygen. This reaction takes place at a temperature of over 570°C

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3 years ago

What is a limiting factor?

alukav5142 [94]

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3 years ago